I keep finding myself interested in this question.

Question

kainui · Answer

This is what's called the finite difference, 
$\Delta f(n) = f(n+1)-f(n)$

So that means:

$\Delta n = 1$

$\Delta n^2 = 2n+1$

$\Delta n^3 = 3n^2+3n+1$

$\cdots$

$\Delta n^k =(n+1)^k-n^k = \sum_{i=0}^{k-1} \binom{k-1}{i} n^i$

Now my question is, what's the general form for the reverse relations:

$n = \frac{1}{2} \Delta n^2 - \frac{1}{2} \Delta n$

$n^2 = \frac{1}{3} \Delta n^3 +\frac{1}{2} \Delta n^2 + \frac{1}{6} \Delta n$

$n^3 = \frac{1}{4} \Delta n^4 + \frac{1}{2} \Delta n^3 + \frac{1}{4} \Delta n^2 $

$\cdots$

$n^k = ?$

isaacsgames · Answer

hi

parthkohli · Answer

interesting - from what you've written so far I could only gather that$$\frac{1}4 = \frac{1}2 - \frac{1}4$$$$\frac{1}6= \frac{1}2-\frac{1}3$$My brain may be constructing stupid patterns from this limited information though, lol

parthkohli · Answer

and also that the first term is $\frac{1}{k+1}\Delta n^{k+1}$ but all of these seem pretty obvious

anonymous · Answer

I think there's a slight typo, shouldn't it be
$$\Delta n^k=\sum_{i=0}^{k-1}\binom ki n^i~~?$$The way you currently have it would make the leading term's coefficient $1$, though it should be $\dbinom k{k-1}=\dbinom k1=k$.

Anyway, interesting question.

kainui · Answer

Ah yeah, should just be $\binom{k}{i}$ my bad.

Also, in case you guys were wondering what possible uses this has, you have already seen this before:

Since the finite difference operator is a linear operator, I can push the constants in and factor it out of both of them like this:

$n = \frac{1}{2} \Delta n^2 - \frac{1}{2} \Delta n = \Delta \left( \frac{1}{2} n^2 - \frac{1}{2}  n \right) = \Delta \left( \frac{n(n-1)}{2} \right)$

So, importantly:

$n= \Delta \left( \frac{n(n-1)}{2} \right)$

Sum over both sides, the right hand side is a telescoping series (I like to think fundamental theorem of calculus):

$$\sum_{n=1}^{k-1} n = \sum_{n=1}^{k-1} \Delta \left( \frac{n(n-1)}{2} \right)  = \frac{k(k-1)}{2}$$

Yay.

ganeshie8 · Answer

this somehow reminds of the good old telescoping thingy :
 $$\sum\limits_{i=1}^n (i+1)^k-
i^k = (n+1)^k-1$$

anonymous · Answer

You're looking to write $n^k$ as a linear combination of the $\Delta n^i$s, right? Something tells me
$$n^k=\frac{\Delta n^{k+1}-\displaystyle\sum_{i=0}^{k-1}\binom {k+1}in^i}{\dbinom{k+1}k}$$isn't satisfactory.